Expectation property of moment generating function
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Let $X$ be a non-negative random variable and define $M_X(t) := mathbb{E}[e^{tX}]$. Suppose that $M_X(t) < infty mbox{ for all } t < alpha$, where $alpha > 0$.
It can be shown that this condition implies $mathbb{E}[X^k]<infty$ for all $k$, i.e. all moments exist.
How can we establish that $mathbb{E}[X^ke^{tX}] < infty$ for all $k$ and $ t < alpha$?
probability probability-theory random-variables moment-generating-functions expected-value
asked Nov 22 at 20:36
jesterII
1,19121226
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up vote
0
down vote
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Let $X$ be a non-negative random variable and define $M_X(t) := mathbb{E}[e^{tX}]$. Suppose that $M_X(t) < infty mbox{ for all } t < alpha$, where $alpha > 0$.
It can be shown that this condition implies $mathbb{E}[X^k]<infty$ for all $k$, i.e. all moments exist.
How can we establish that $mathbb{E}[X^ke^{tX}] < infty$ for all $k$ and $ t < alpha$?
probability probability-theory random-variables moment-generating-functions expected-value
asked Nov 22 at 20:36
jesterII
1,19121226
My idea is to use tail property. Any expectation is integral (supposedly either in Stieltjes or in Lebesgue sense.) So you can divide your last expectation into the parts $[0 , epsilon]$ and $[epsilon , infty)$. And then analyse the convergence properties. (In the first region it is for sure bounded), In tail regions it can be done intuitively. $e^{x}$ is faster then $dF_{X}(x)$ in tail, so $dF_{X}(x)= o(e^{-x})$, such that for no $k$, $e^{x}*x^{k}$ can be faster $dF_{X}(x)$.(This is acc. proof, when you have continuous random variable, but I'm not sure with general case(can have jumps)).
– kolobokish
Nov 22 at 20:51
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ be a non-negative random variable and define $M_X(t) := mathbb{E}[e^{tX}]$. Suppose that $M_X(t) < infty mbox{ for all } t < alpha$, where $alpha > 0$.
It can be shown that this condition implies $mathbb{E}[X^k]<infty$ for all $k$, i.e. all moments exist.
How can we establish that $mathbb{E}[X^ke^{tX}] < infty$ for all $k$ and $ t < alpha$?
probability probability-theory random-variables moment-generating-functions expected-value
asked Nov 22 at 20:36
jesterII
1,19121226
Let $X$ be a non-negative random variable and define $M_X(t) := mathbb{E}[e^{tX}]$. Suppose that $M_X(t) < infty mbox{ for all } t < alpha$, where $alpha > 0$.
It can be shown that this condition implies $mathbb{E}[X^k]<infty$ for all $k$, i.e. all moments exist.
How can we establish that $mathbb{E}[X^ke^{tX}] < infty$ for all $k$ and $ t < alpha$?
probability probability-theory random-variables moment-generating-functions expected-value
probability probability-theory random-variables moment-generating-functions expected-value
asked Nov 22 at 20:36
jesterII
1,19121226
asked Nov 22 at 20:36
jesterII
1,19121226
asked Nov 22 at 20:36
jesterII
1,19121226
asked Nov 22 at 20:36
jesterII
1,19121226
asked Nov 22 at 20:36
jesterII
1,19121226
1,19121226
My idea is to use tail property. Any expectation is integral (supposedly either in Stieltjes or in Lebesgue sense.) So you can divide your last expectation into the parts $[0 , epsilon]$ and $[epsilon , infty)$. And then analyse the convergence properties. (In the first region it is for sure bounded), In tail regions it can be done intuitively. $e^{x}$ is faster then $dF_{X}(x)$ in tail, so $dF_{X}(x)= o(e^{-x})$, such that for no $k$, $e^{x}*x^{k}$ can be faster $dF_{X}(x)$.(This is acc. proof, when you have continuous random variable, but I'm not sure with general case(can have jumps)).
– kolobokish
Nov 22 at 20:51
add a comment |
My idea is to use tail property. Any expectation is integral (supposedly either in Stieltjes or in Lebesgue sense.) So you can divide your last expectation into the parts $[0 , epsilon]$ and $[epsilon , infty)$. And then analyse the convergence properties. (In the first region it is for sure bounded), In tail regions it can be done intuitively. $e^{x}$ is faster then $dF_{X}(x)$ in tail, so $dF_{X}(x)= o(e^{-x})$, such that for no $k$, $e^{x}*x^{k}$ can be faster $dF_{X}(x)$.(This is acc. proof, when you have continuous random variable, but I'm not sure with general case(can have jumps)).
– kolobokish
Nov 22 at 20:51
My idea is to use tail property. Any expectation is integral (supposedly either in Stieltjes or in Lebesgue sense.) So you can divide your last expectation into the parts $[0 , epsilon]$ and $[epsilon , infty)$. And then analyse the convergence properties. (In the first region it is for sure bounded), In tail regions it can be done intuitively. $e^{x}$ is faster then $dF_{X}(x)$ in tail, so $dF_{X}(x)= o(e^{-x})$, such that for no $k$, $e^{x}*x^{k}$ can be faster $dF_{X}(x)$.(This is acc. proof, when you have continuous random variable, but I'm not sure with general case(can have jumps)).
– kolobokish
Nov 22 at 20:51
My idea is to use tail property. Any expectation is integral (supposedly either in Stieltjes or in Lebesgue sense.) So you can divide your last expectation into the parts $[0 , epsilon]$ and $[epsilon , infty)$. And then analyse the convergence properties. (In the first region it is for sure bounded), In tail regions it can be done intuitively. $e^{x}$ is faster then $dF_{X}(x)$ in tail, so $dF_{X}(x)= o(e^{-x})$, such that for no $k$, $e^{x}*x^{k}$ can be faster $dF_{X}(x)$.(This is acc. proof, when you have continuous random variable, but I'm not sure with general case(can have jumps)).
– kolobokish
Nov 22 at 20:51
add a comment |
1 Answer
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Let $t<t'<alpha$. Then $e^{(t'-t)x} geq frac {(t'-t)^{k}x^{k}} {k!}$ so $x^{k}e^{tx} leq frac {e^{t'x} k!} {(t'-t)^{k}}$. Hence $EX^{k}e^{tX}leq M(t') frac { k!} {(t'-t)^{k}}<infty$.
answered Nov 22 at 23:50
Kavi Rama Murthy
47.3k31854
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $t<t'<alpha$. Then $e^{(t'-t)x} geq frac {(t'-t)^{k}x^{k}} {k!}$ so $x^{k}e^{tx} leq frac {e^{t'x} k!} {(t'-t)^{k}}$. Hence $EX^{k}e^{tX}leq M(t') frac { k!} {(t'-t)^{k}}<infty$.
answered Nov 22 at 23:50
Kavi Rama Murthy
47.3k31854
add a comment |
up vote
2
down vote
accepted
Let $t<t'<alpha$. Then $e^{(t'-t)x} geq frac {(t'-t)^{k}x^{k}} {k!}$ so $x^{k}e^{tx} leq frac {e^{t'x} k!} {(t'-t)^{k}}$. Hence $EX^{k}e^{tX}leq M(t') frac { k!} {(t'-t)^{k}}<infty$.
answered Nov 22 at 23:50
Kavi Rama Murthy
47.3k31854
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $t<t'<alpha$. Then $e^{(t'-t)x} geq frac {(t'-t)^{k}x^{k}} {k!}$ so $x^{k}e^{tx} leq frac {e^{t'x} k!} {(t'-t)^{k}}$. Hence $EX^{k}e^{tX}leq M(t') frac { k!} {(t'-t)^{k}}<infty$.
answered Nov 22 at 23:50
Kavi Rama Murthy
47.3k31854
Let $t<t'<alpha$. Then $e^{(t'-t)x} geq frac {(t'-t)^{k}x^{k}} {k!}$ so $x^{k}e^{tx} leq frac {e^{t'x} k!} {(t'-t)^{k}}$. Hence $EX^{k}e^{tX}leq M(t') frac { k!} {(t'-t)^{k}}<infty$.
answered Nov 22 at 23:50
Kavi Rama Murthy
47.3k31854
answered Nov 22 at 23:50
Kavi Rama Murthy
47.3k31854
answered Nov 22 at 23:50
Kavi Rama Murthy
47.3k31854
answered Nov 22 at 23:50
Kavi Rama Murthy
47.3k31854
47.3k31854
add a comment |
add a comment |
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My idea is to use tail property. Any expectation is integral (supposedly either in Stieltjes or in Lebesgue sense.) So you can divide your last expectation into the parts $[0 , epsilon]$ and $[epsilon , infty)$. And then analyse the convergence properties. (In the first region it is for sure bounded), In tail regions it can be done intuitively. $e^{x}$ is faster then $dF_{X}(x)$ in tail, so $dF_{X}(x)= o(e^{-x})$, such that for no $k$, $e^{x}*x^{k}$ can be faster $dF_{X}(x)$.(This is acc. proof, when you have continuous random variable, but I'm not sure with general case(can have jumps)).
– kolobokish
Nov 22 at 20:51